If the gravitational force between two objects were proportional to $ 1/R $ (and not as $ 1/R^2 $ ), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to
- R
- $ R^0$ (independent of R)
- $1/R^2$
- $1/R$
The Correct Option is B
Solution and Explanation
gravitational force (F) $\frac{ GMm}{R^2} =\frac{ GMm}{R}$ (where
$R^2 \rightarrow R ) $ . Since $ \frac{ mv^2}{R} = \frac{ GMm}{R} $, therefore $ v= \sqrt{GM }$.
Thus velocity v is independent of R.
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Concepts Used:
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In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].